AMC12 2018 B

Answer (C): Let $\{x\}=x-[x]$ denote the fractional part of $x$. Then $0\le \{x\}<1$. The given equation is equivalent to $x^2=10{,}000\{x\}$, that is, $\dfrac{x^2}{10{,}000}=\{x\}.$ Therefore if $x$ satisfies the equation, then $0\le \dfrac{x^2}{10{,}000}<1.$ This implies that $x^2<10{,}000$, so $-100<x<100$. The figure shows a sketch of the graphs of $f(x)=\dfrac{x^2}{10{,}000}\quad$ and $\quad g(x)=\{x\}$ for $-100<x<100$ on the same coordinate axes. The graph of $g$ consists of the 200 half-open line segments with slope 1 connecting the points $(k,0)$ and $(k+1,1)$ for $k=-100,-99,\ldots,98,99$. (The endpoints of these intervals that lie on the $x$-axis are part of the graph, but the endpoints with $y$-coordinate 1 are not.) It is clear that there is one intersection point for $x$ lying in each of the intervals $[-100,-99)$, $[-99,-98]$, $[-98,-97)$, $\ldots$, $[-1,0)$, $[0,1)$, $[1,2)$, $\ldots$, $[97,98)$, $[98,99)$ but no others. Thus the equation has 199 solutions.